Question

Definite integration:

$\boxed{\displaystyle \frac{\int\limits^\pi_0 (1+\cos \theta) (1-\cos \theta) \,d\theta}{\int\limits_0^\pi (\sqrt{1+\cos\theta})(\sqrt{1-\cos \theta})\, d\theta}}$

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\displaystyle \dfrac{\displaystyle\int ^\pi_0 (1+\cos \theta) (1-\cos \theta) \,d\theta}{\displaystyle\int ^\pi_0 (\sqrt{1+\cos\theta})(\sqrt{1-\cos \theta})\, d\theta}$

Now, Consider

$\rm \: \displaystyle\int ^\pi_0(1 + cos\theta)(1 - cos\theta) \: d\theta$

$\rm \: \displaystyle\int ^\pi_0(1 - {cos}^{2}\theta) \: d\theta$

$\rm \: \displaystyle\int ^\pi_0({sin}^{2}\theta) \: d\theta$

$\rm \: = \: \dfrac{1}{2} \displaystyle\int ^\pi_0 \: 2{sin}^{2}\theta \: d\theta$

$\rm \: = \: \dfrac{1}{2} \displaystyle\int ^\pi_0 \: (1 - cos2\theta )\: d\theta$

$\rm \: = \: \dfrac{1}{2}\bigg[\theta - \dfrac{sin2\theta}{2} \bigg] ^\pi_0$

$\rm \: = \: \dfrac{1}{2}\bigg[(\pi - 0) - \dfrac{sin2\pi - sin0}{2} \bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[\pi - \dfrac{0 - 0}{2} \bigg]$

$\rm \: = \: \dfrac{1}{2}\bigg[\pi - 0 \bigg]$

$\rm \: = \: \dfrac{\pi}{2}$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int ^\pi_0(1 + cos\theta)(1 - cos\theta) \: d\theta \: = \: \frac{\pi}{2} \: }}$

Now, Consider

$\rm \: \displaystyle\int ^\pi_0 \: \sqrt{(1 + cos\theta)} \: \sqrt{(1 - cos\theta)} \: d\theta$

$\rm \: = \: \displaystyle\int ^\pi_0 \: \sqrt{(1 + cos\theta)(1 - cos\theta)} \: d\theta$

$\rm \: = \: \displaystyle\int ^\pi_0 \: \sqrt{1 - cos^{2} \theta} \: d\theta$

$\rm \: = \: \displaystyle\int ^\pi_0 \: \sqrt{sin^{2} \theta} \: d\theta$

$\rm \: = \: \displaystyle\int ^\pi_0 \: |sin\theta| \: d\theta$

As we know, sinx > 0 in first and second quadrant.

So, using this

$\rm \: = \: \displaystyle\int ^\pi_0 \: sin\theta \: d\theta$

$\rm \: = \: - \bigg(cos\theta \bigg) ^\pi_0 \:$

$\rm \: = \: - \bigg(cos\pi - cos0 \bigg)$

$\rm \: = \: - \bigg( - 1- 1\bigg)$

$\rm \: = \: - \bigg( -2\bigg)$

$\rm \: = \: 2$

So,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int ^\pi_0 \: \sqrt{(1 + cos\theta)} \: \sqrt{(1 - cos\theta)} \: d\theta = 2 \: }}$

Now, Consider the given integral,

$\displaystyle \dfrac{\displaystyle\int ^\pi_0 (1+\cos \theta) (1-\cos \theta) \,d\theta}{\displaystyle\int ^\pi_0 (\sqrt{1+\cos\theta})(\sqrt{1-\cos \theta})\, d\theta}$

So, on substituting the values, evaluated above, we get

$\rm \: = \: \dfrac{\dfrac{\pi}{2} }{ \: \: 2 \: \: }$

$\rm \: = \: \dfrac{\pi}{4}$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle \dfrac{\displaystyle\int ^\pi_0 (1+\cos \theta) (1-\cos \theta) \,d\theta}{\displaystyle\int ^\pi_0 (\sqrt{1+\cos\theta})(\sqrt{1-\cos \theta})\, d\theta} = \frac{\pi}{4}}}$

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FORMULAE USED

$\rm \: {cos}^{2}x + {sin}^{2}x = 1$

$\rm \: \sqrt{ {x}^{2} } = |x|$

$\displaystyle\int\rm cosx \: dx = sinx \: + \: c$

$\displaystyle\int\rm sinx \: dx = - \: cosx \: + \: c$

$\displaystyle\int\rm k \: dx = kx \: + \: c$

$\rm \: sin \: n\pi \: = \: 0 \: where \: n \: is \: integer$

$\rm \: cos \: n\pi \: = \: {( - 1)}^{n} \: where \: n \: is \: integer$